I have following term:
$$\frac{\frac{3^{2k+2}}{(2k+2)!}}{\frac{2^{2k}}{(2k)!}}=\frac{3^{2k+2}\cdot(2k)!}{(2k+2)! \cdot 3^{2k}}=9\cdot\frac{(2k)!}{(2k+2)!}$$
I know that you can simplify even further to $$\frac{9}{(2k+2)(2k+1)}$$ but I do not get the intermediate steps.
I know that $$(n+1)! = n!(n+1) = (n-1)! \cdot (n+1) \cdot n$$
but I still do not know how to apply this on terms different from $(n+1)$ (e.g. $(2n+2)$)?
You're right it can be simplified
$$9\frac{(2k)!}{(2k+2)(2k+1)(2k)!} = \frac{9}{(2k+2)(2k+1)}$$